Question: Multiply the following complex numbers: $({-3-2i}) \cdot ({-2+i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-3-2i}) \cdot ({-2+i}) = $ $ ({-3} \cdot {-2}) + ({-3} \cdot {1}i) + ({-2}i \cdot {-2}) + ({-2}i \cdot {1}i) $ Then simplify the terms: $ (6) + (-3i) + (4i) + (-2 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 6 + (-3 + 4)i - 2i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 6 + (-3 + 4)i - (-2) $ The result is simplified: $ (6 + 2) + (1i) = 8+i $